"... it happens to cause the core to ring at the 4th (of 3.58mHz) and the
MOSFET starts up as a parametric pump. I adjust the phase with the FE beads
and the circuit is totally AC under its own power. Complicated to view the
thing, but with care is possible. Has a field around it that is quite large.

Forum

If you apply 2 Volts peak to peak via a 10 KOhm resistor (if it is high frequency, the 10 pF cap mentioned,might just be conducting..so I neglected it..) so the input power is around 400 microWatts. The output power is 2.73 Volts x 25 miliamps = 68.25 milliWatts. So ouput / input is around 170. (Stefan Hartmann; Oct. 11, 2007)

comments on video 2 by PM:

This circuit is similar to a very common circuit used by thousands of people everyday to
power the flash lamps in disposable cameras.

The equation Ronald Stiffler is using for energy in the capacitor is based on the voltage of
an open circuit capacitor. The measurements seem to be of the voltage being applied to
the capacitor in the circuit. Depending on the circuit, these can be quite different.
Therefore his capacitor energy calculations are probably incorrect.

The readings shown on the power supply are 12 volts and 0.120 amps. Power in watts is
volts times amps. Power is an instantaneous reading. Energy is power times seconds.
Energy is an average reading over time. The power going into the circuit is 12 volts x
0.120 amps = 1.44 watts constantly. The energy going into the circuit is the average
power per second. Since the power going into the circuit is constant, the average value
will be the same as the constant value. Thus the energy going into the circuit is 1.44
watts-seconds where 1.44 watts is the average value of the power during each second.
This energy is then converted to a higher voltage by use of the transformer (apparently to
about 114 volts). Since energy is conserved, the energy coming out of the transformer is
also 1.44 watt-seconds (with some losses due to heat). This energy can be stored in a
capacitor and removed either slowly or quickly as desired. In this case all of the energy
added to the capacitor in each second is then removed at the end of the second (if the light
is pulsed once per second as stated in the video). Thus an energy of 1.44 watt-seconds is
added to and then removed from the capacitor each second. It is not given, but the bulb
appears to be on for about 1/10 of each second. Assuming this value, then the average
watts going into the bulb during that 1/10 of a second must be 10 x 1.44 = 14.4 watts in
order for the energy to be 1.44 watt-seconds (with 0 watts going into bulb when it is off).
Remember, energy is the average value of the instantaneous power in watts that is applied
during the second. In this case, to get the average power put into the bulb the power in
each 1/10 of the second is summed and then divided by 10 > 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0
+ 0 + 14.4 = 14.4 and 14.4/10 = 1.44. Since the bulb is a 15 watt bulb, it will be bright
during the 1/10 second it is consuming 14.4 watts. Thus the circuit performance can easily
be explained without “cold electricity? or “over unity?.

Note that the primary of the transformer is being fed with an RF frequency signal. It is
unclear if this is deliberate or if it is a parasitic oscillation. The open lead on the secondary
of the transformer is not actually unconnected at RF frequencies but is connected to other
parts of the circuit by the parasitic capacitance of the lead. Same for the neon bulbs.